Electrical Circuit Question

1. Jan 30, 2013

GreenPrint

http://img15.imageshack.us/img15/771/captureuce.png [Broken]
1. The problem statement, all variables and given/known data

Determine if the circuit is valid or invalid.

2. Relevant equations

3. The attempt at a solution

Please note which node I used as the ground node and which node I have labeled $V_{1}$. By the node voltage method

$\frac{V_{1}}{8} + 10 - 2 + \frac{V_{1} - 10}{5} = 0$
$\frac{13V_{1}}{40} + 6 = 0$
$\frac{13V_{1}}{40} = -6$
$V_{1} = -\frac{240}{13} V$

$I_{1} = \frac{V_{1}}{8} = -\frac{240}{13*8} = -\frac{30}{13} A$
$I_{2} = \frac{V_{1}-10}{5} = \frac{\frac{-240}{13}-10}{5} = -\frac{74}{13}$
$I_{3} = 10 + I_{2} = 10 + {-\frac{74}{13}} = \frac{56}{13}$

$V = IR$
$P = IV$
$P = I^{2}R$

$P_{8Ω} = I^{2}R = (-\frac{30}{13})^{2}8 = \frac{7200}{169}$
$P_{5Ω} = I^{2}R = (-\frac{74}{13})^{2}5 = \frac{27380}{169}$
$P_{2Ω} = I^{2}R = (\frac{56}{13})^{2}2 = \frac{6272}{169}$
$P_{10V} = IV = \frac{56}{13}10 = \frac{560}{13}$
$P_{2A} = \frac{240}{13}2 = IV = \frac{480}{13}$
$P_{10A} = IV = 10{-\frac{240}{13}-\frac{74}{13}} = \frac{3140}{13}$

These don't add up to zero. I'm kind of confused as to when you are supposed to use a negative sign or not. I used a negative sign in the last calculation because the current is going from lower potential to higher potential. Thanks for nay help you can provide.

Last edited by a moderator: May 6, 2017
2. Jan 30, 2013

Staff: Mentor

In your first equation, the final term $\frac{V_1 - 10}{5}$ is not correct. The point where the 5Ω resistor terminates is not at 10V above reference since there's a series resistor of 2Ω in the path. You have two nodes to deal with.

Last edited: Jan 30, 2013
3. Jan 30, 2013

GreenPrint

Hi gneill,

Why can't I count the node where the 10 volt source, 5 ohm resistor, and 10 Amp source meet to be 10 volts?

4. Jan 30, 2013

Staff: Mentor

You could if the reference node was at the bottom of the 10V source. But it isn't. There's a 2Ω resistor in the way.

5. Jan 30, 2013

GreenPrint

Ok thanks I forgot about that resistor.

6. Jan 30, 2013

GreenPrint

http://img59.imageshack.us/img59/1566/capturekwm.png [Broken]

I believe the picture in this post is accurate. However how would I calculated the power dissipated in the 2A source?

I know I use

P = IV
P = 2(0-(-19.2))
or
P = 2(-19.2-0)

which one is correct?

Thanks

Last edited by a moderator: May 6, 2017
7. Jan 30, 2013

Staff: Mentor

Your node voltages and currents don't look right. Better recheck your calculations.

If the potential across a current source is negative (if it's driving a current from a higher potential to a lower potential), then it's absorbing energy.

8. Jan 31, 2013

GreenPrint

What exactly dosen't look right?

9. Jan 31, 2013

Staff: Mentor

All the voltages and currents that you've added to the drawing.